Question: What is the area of the region bound by the graphs of $f(x)=-x^2$, $g(x)=3x-10$, and $x=0$ in quadrant $\text{IV}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $2$ (Choice B) B $\dfrac{16}{3}$ (Choice C) C $\dfrac{8}{3}$ (Choice D) D $\dfrac{34}{3}$
Explanation: Visualizing the area We sketch the graphs of $f$ and $g$ first. ${1}$ ${2}$ ${3}$ ${\llap{-}4}$ ${\llap{-}6}$ ${\llap{-}8}$ ${\llap{-}10}$ ${\llap{-}12}$ ${\llap{-}14}$ $f$ $g$ $y$ $x$ From the graph, it appears that $f(x)\ge g(x)$ between the points where the graphs intersect. From this we are looking to evaluate: $ \int_{0}^{b}\left( f(x)-g(x) \right)\,dx$ where $b$ is the $x$ -coordinate of the point of intersection in quadrant $\text{IV}$. Finding the $x$ -coordinates of the intersection points We can find the $x$ -coordinate of each point of intersection by setting the functions equal to each other and solving the resulting equation. $\begin{aligned} f(x)&=g(x) \\\\ -x^2&=3x-10\\\\ 0 &= x^2+3x-10 \\\\ 0 &= (x+5)(x-2) \end{aligned}$ The graphs intersect where $x=-5$ and $x=2$. Because we only want the part of the graph in quadrant $\text{I}V$, we will use $x=2$ as our upper bound. Setting up the definite integral Thus, the area of the shaded region pictured above is given by: $\begin{aligned} &\phantom{=} \int_{0}^{2}\left(-x^2-(3x-10)\right)\,dx \\\\ &= \int_{0}^{2}\left(-x^2-3x+10\right)\,dx \end{aligned}$ Evaluating the definite integral $\begin{aligned} &\phantom{=} \int_{0}^{2} \left( -x^2- 3x+10 \right) \,dx \\\\ &= -\dfrac{x^3}{3}-\dfrac{3x^2}{2}+10x~\Bigg|_{0}^{2} \\\\ &= \left( -\dfrac{8}{3}-6+20 \right) - \left( 0-0+0 \right) \\\\ &= \dfrac{34}{3} \end{aligned}$ Answer The area is $\dfrac{34}{3}$ square units.